Problem: On the way to school, Karen became fixated on the puzzle game on her phone!

<image>

The game is played as follows. In each level, you have a grid with n rows and m columns. Each cell originally contains the number 0.

One move consists of choosing one row or column, and adding 1 to all of the cells in that row or column.

To win the level, after all the moves, the number in the cell at the i-th row and j-th column should be equal to gi, j.

Karen is stuck on one level, and wants to know a way to beat this level using the minimum number of moves. Please, help her with this task!

Input

The first line of input contains two integers, n and m (1 ≤ n, m ≤ 100), the number of rows and the number of columns in the grid, respectively.

The next n lines each contain m integers. In particular, the j-th integer in the i-th of these rows contains gi, j (0 ≤ gi, j ≤ 500).

Output

If there is an error and it is actually not possible to beat the level, output a single integer -1.

Otherwise, on the first line, output a single integer k, the minimum number of moves necessary to beat the level.

The next k lines should each contain one of the following, describing the moves in the order they must be done:

  * row x, (1 ≤ x ≤ n) describing a move of the form "choose the x-th row". 
  * col x, (1 ≤ x ≤ m) describing a move of the form "choose the x-th column". 



If there are multiple optimal solutions, output any one of them.

Examples

Input

3 5
2 2 2 3 2
0 0 0 1 0
1 1 1 2 1


Output

4
row 1
row 1
col 4
row 3


Input

3 3
0 0 0
0 1 0
0 0 0


Output

-1


Input

3 3
1 1 1
1 1 1
1 1 1


Output

3
row 1
row 2
row 3

Note

In the first test case, Karen has a grid with 3 rows and 5 columns. She can perform the following 4 moves to beat the level:

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In the second test case, Karen has a grid with 3 rows and 3 columns. It is clear that it is impossible to beat the level; performing any move will create three 1s on the grid, but it is required to only have one 1 in the center.

In the third test case, Karen has a grid with 3 rows and 3 columns. She can perform the following 3 moves to beat the level:

<image>

Note that this is not the only solution; another solution, among others, is col 1, col 2, col 3.

Solution (in JAVA):

Solution:
import javafx.util.Pair;

import javax.swing.text.html.HTMLDocument;

import static java.lang.Double.parseDouble;
import static java.lang.Integer.max;
import static java.lang.Integer.min;
import static java.lang.Integer.parseInt;
import static java.lang.Long.parseLong;
import static java.lang.StrictMath.abs;
import static java.lang.System.err;
import static java.lang.System.exit;
import static java.util.Arrays.sort;
import java.io.*;
import java.lang.reflect.Array;
import java.math.BigInteger;
import java.util.*;


public class A {

    static BufferedReader in;
    static PrintWriter out;
    static StringTokenizer tok;


    static void solve() throws Exception {
        int n = nextInt() , m = nextInt();
        int a[][] = new int[n][m];
        int aa[][] = new int[n][m];
        boolean ok1 = true, ok2 = true;
        ArrayList<Integer> row1 = new ArrayList<>();
        ArrayList<Integer> column1 = new ArrayList<>();
        for(int i=0;i<n;i++) for(int j=0;j<m;j++) {
            a[i][j] = nextInt();
            aa[i][j] = a[i][j];
        }
        // 1
        for(int i=0;i<n;i++) {
            int mn = 500;
            for(int j=0;j<m;j++) mn = min(mn,a[i][j]);
            for(int j=0;j<m;j++) a[i][j] -= mn;
            for(int j=0;j<mn;j++) row1.add(i+1);
        }
        for(int j=0;j<m;j++) {
            for(int i=1;i<n;i++) if(a[i][j] != a[i-1][j]) {
                ok1 = false;
            }
            for(int i=0;i<a[0][j];i++) column1.add(j+1);
        }
        // 2

        ArrayList<Integer> row2 = new ArrayList<>();
        ArrayList<Integer> column2 = new ArrayList<>();
        for(int j=0;j<m;j++) {
            int mn = 500;
            for(int i=0;i<n;i++) mn = min(mn,aa[i][j]);
            for(int i=0;i<n;i++) aa[i][j] -= mn;
            for(int i=0;i<mn;i++) column2.add(j+1);
        }
        for(int i=0;i<n;i++) {
            for(int j=1;j<m;j++) if(aa[i][j] != aa[i][j-1]) {
                ok2 = false;
            }
            for(int j=0;j<aa[i][0];j++) row2.add(i+1);
        }
        // result
        if(ok1 && ok2) {
            if(row1.size() + column1.size() <= row2.size() + column2.size()) {
                out.println(row1.size() + column1.size());
                for(int x:row1) out.println("row " + x);
                for(int x:column1) out.println("col " + x);
            } else {
                out.println(row2.size() + column2.size());
                for(int x:row2) out.println("row " + x);
                for(int x:column2) out.println("col " + x);
            }
        }
        else if(ok1) {
            out.println(row1.size() + column1.size());
            for(int x:row1) out.println("row " + x);
            for(int x:column1) out.println("col " + x);
        }
        else if(ok2) {
            out.println(row2.size() + column2.size());
            for(int x:row2) out.println("row " + x);
            for(int x:column2) out.println("col " + x);
        } else out.print(-1);

    }

    public static int lower_bound(ArrayList<Integer> a, int v) {
        int low = -1, high = a.size();
        while(high-low > 1){
            int h = high+low>>>1;
            if(a.get(h) >= v){
                high = h;
            }else{
                low = h;
            }
        }
        return high;
    }

    public static int upper_bound(long[] a, int v) {
        int low = -1, high = a.length;
        while(high-low > 1){
            int h = high+low>>>1;
            if(a[h] > v){
                high = h;
            }else{
                low = h;
            }
        }
        return high;
    }

    static int gcd(int a, int b) { return b==0 ? a : gcd(b, a%b); }


    static int nextInt() throws IOException {
        return parseInt(next());
    }

    static long nextLong() throws IOException {
        return parseLong(next());
    }

    static double nextDouble() throws IOException {
        return parseDouble(next());
    }

    static String next() throws IOException {
        while (tok == null || !tok.hasMoreTokens()) {
            tok = new StringTokenizer(in.readLine());
        }
        return tok.nextToken();
    }

    public static void main(String[] args) {
        try {
            in = new BufferedReader(new InputStreamReader(System.in));
            out = new PrintWriter(new OutputStreamWriter(System.out));
            solve();
            long lEndTime = System.currentTimeMillis();
            //System.out.println("Elapsed time in seconds: " + (double)(lEndTime - lStartTime) / 1000.0);
            in.close();
            out.close();
        } catch (Throwable e) {
            e.printStackTrace();
            exit(1);
        }
    }
}